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In each of the following exercises, solve the given inequality.$$x^{2}-2 x-1 \geq 0$$

$$x \leq 1-\sqrt{2} \text { or } x \geq 1+\sqrt{2}$$

Algebra

Chapter 0

Reviewing the Basics

Section 5

Solving Non-Linear Inequalities

Equations and Inequalities

Missouri State University

McMaster University

Baylor University

Lectures

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In each of the following e…

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Solve each inequality alge…

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Solve each inequality.…

All right, we've got a question here. X squared minus two X minus. One is greater than you could do. Zero. We want a factor this out. First, we can see that it's not a factor. A polynomial that could be factored easily. And we would have to use the quadratic formula. When you go ahead and plug it into your quadratic formula, you get two values for X X is you could excuse me. X is equal to one plus route to an X is equal to one minus root, too. From there gonna take you're gonna create a number line that goes from negative infinity to infinity and you'll plot those values on your number one. Okay, you're gonna have them be solid points because you also have the equal to as long as it's equal to these values, you will satisfy the inequality. Then you want to determine if any of these three regions here also will be able to satisfy this inequality. So we'll pick a value between one plus route to an infinity first, let's just choose three. Could be any number between those that within that region, and we'll get here. We'll get nine minus six is 33 Minus one is to that right When doing that, right, guys? Yes. To so a positive two is greater than equal to zero. Therefore, this does satisfy inequality. We'll pick a value in between those two. We could pick zero zero minus zero minus one. It's negative. One negative one. It's certainly not greater than or equal to zero. So we'll say, Well, it doesn't exist here. Finally, will pick a value between one minus route to a negative infinity so we could choose negative three squared multiplied by two times. Excuse me. Subtracting two times negative. Three minus one, Which would give us a negative six plus minus one. Excuse me. So what have nine minus and negative? Six. Which you make a positive 15 minus one, which would be 14. That's a positive value. And that's greater than zero. So exists it exists within that region as well. And then we're gonna write down our X domain here, and we'll say, Well, whenever X is less than or equal to one minus root too. Okay. And whenever X is greater than or equal to one plus two, the inequality is satisfied. All right, well, I hope that clarifies the question there. Thank you so much for watching

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